ECN Forum Forums Technical Discussion Electrical Theory and Applications Calculations

JBD,
>>>My brain hurts<<<

No pain, no gain....

It wasn't my description, it was originaly posted at Mike Holts site. I just figured we could throw it around and see what we come up with...

I don't think it matters if they are 120V loads or 240V loads on the A,C phases. I think that is what threw me off.

I used the sq root of A sqr'd formula to get the nuetral current.(don't know how to post it) I am not sure how Radar and Paul get their nuetral current answer though....

Now get them brain cells moving and let's see some answers...

Dnk...

[This message has been edited by Dnkldorf (edited 02-10-2006).]

Guys, Gerald Newton's form is up and running again and I think he need a little help getting more posters. His form got hit real hard with spam about a year or so ago and he got tired of fighting it and shut it down. It's been back up for a couple of months an has been getting very little activity. The main site is: http://electrician2.com/
The forum: http://electrician2.com/elwwwboard/wwwboard.html
Gerald is an Alaska guy.

Dnk - you know that in a household 120/240V service, the neutral carries only the unbalanced portion of the line current, the difference between L1 current and L2 current. In your example here, ignoring that fact that ØB even exists for the moment, the same thing is true. On ØC, 34KVA of the 41KVA load is balanced by the 34KVA on ØA, and the resulting current flows between ØC and ØA, not involving the neutral.

The unbalanced part of the load, 7KVA, is seen between ØC and the neutral. 7,000VA divided by 120V is approx equal to 58 amps. That is to say, the unbalanced part is 58 amps and flows between ØC and neutral.

The SINGLE PHASE component of the loads are sumarized thus (Kirchoff's Current Law):
ØA = 284 amps
Neutral = 58 amps
ØC = 342 amps (284 + 58)

I'm not sure I'm any good at describing this stuff, hope this helps.

Radar

Radar, I got you, but that is a split phase, 180 degree phase system. The above, would be a 3 phase 120 degree phase system. no?

In a three phase system(I'm gonna try)


N= Sq root of (A^+B^+C^)-(AB+AC+CB)

^=squared

Is this the wrong formula?

Did you get the same phase currents as I did?


Dnk..

Dnk,

You are spot on. In a two leg _single_ phase system, you need two equal currents on each leg to balance to zero on the neutral. In a three leg _three_ phase system, you need _three_ equal currents on each leg to balance to zero on the neutral (I am assuming similar power factor and no harmonics).

If you had 10A on phase A, 10A on phase B, and 0A on phase C, then you would see _10A_ on the neutral.

Oh, for future reference, '^' is often use to mean 'raised to the power', so squared would be '^2' and cubed would be '^3'.

-Jon

Dnk - I think that although we are talking about a 3 phase system overall, the 120/240V 1Ø 3W between L1 & L3 is, in essence, single phase power (disregarding L2 and the other 2 transformers). Consider that single phase resi power is frequently also derived from 2 legs of a 3 phase system.

Anyway, the overall voltage between L1 and L3 (ØA & ØC) is 240V Single Phase with a grounded neutral connection at the centerpoint. IMHO there is little (if any) difference theory wise between this and a resi system.

I see what you're saying, if you connected the 3 line wires to a multi-trace O-scope, each sine wave would be 120º apart from the others. However, if you just look at L1 to L3 on a scope, you'd see a single 240VRMS sine wave, as there is but a single potential between L1 & L3.

Not sure I'm making sense here, please pass the gin. It's been a long week.

Radar

Your right Radar, however....

On a residential "split phase" 240v system the phase voltages are 180 out of phase, and so are the nuetral currents. The nuetral currents, when they meet at a central point, cancel each other out, because they are 180 apart.

Now on a 3 phase system, if we use 2 of the phases, this is a "single phase" 240V circuit. These currents are 120 degrees apart, so when the meet at a central point, they do not completely cancel, like in a "split phase" system.


The nuetral calculations are different...
And to make matters worse, Redsy just threw a hole in my calculations with a new thought, maybe he'll chime in.....

It has been a long week, and I have a weekend of shoveling the white stuff ahead of me...

So, who has the answers to the above problem? It's killing me.....

Or are you guys keeping me in suspense, just to be cruel and give me more grey hair?


Dnk....



[This message has been edited by Dnkldorf (edited 02-10-2006).]

Ignoring the multiple ambiguities as pointed out by JBD, and taking the wording at face value, I'd say that the 80 kva per phase is simply 80 kva, 3-phase. Just like 50 amps per phase is simply 50 amps, 3-phase.
Therefore, the amperages would appear to be
80,000/(240x1.73) = 192 amps
Then, taking the single phase numbers at face value, I'd assume L-N connections because the problem specifies individual phases. This means that phase A would have an additional 283 amps, and phase C, an additional 341 amps.
So...

Phase A = 192 + 283 = 475 amps
Phase B = 192 amps
Phase C = 192 + 343 = 535 amps


Then again, according to some hard-to-follow explanation to a similar sounding problem in "Delmar's Standard Textbook of Electricity" these numbers are probably wrong.




[This message has been edited by Redsy (edited 02-10-2006).]

I'm alergic to that white stuff, makes me start to shake and shivver. Blimey, that's why I live herre I guess.

I'm having a hard time seeing any real difference between the 120V components of a 120/240 delta service and a resi service right next door that's taken off the same primary. In the resi you just have some missing components (the other 2 transformers). Kind of a new definition of a really really open delta.

Let's back up a little and change the original situation just a little. Supposing we had 34KVA on ØA and the same on ØC (instead of 41KVA). That would essentially equate to 68KVA across a 240V potential and render the neutral connection an anchor or ground reference point without any current. Current would flow in A and out C, or visa versa at any given instant of time, neutral current would be zero, A-C current would be 284 amps.

I'm still seeing a single 240V potential between 2 points (ØA & ØC) with a grounded center point, nothing more. You've got a single transformer secondary we're dealing with here, the other transformers connected to L2 (ØB) are of no consequence to the 120/240 part of this. IFF (that's if and only if) any of the 120V current also flowed thru the other transformers to L2 (B), then we'd have phase angles and some trig to work out, but that's not the case. All 120V current flow is limited to the one transformer secondary between L1 & L2. We're splitting that in half, so it appears to be a 180º split.

True enough the three phase currents are flowing all around the transformer connections following the 3 distinct sine waves, but the single phase 120/240 leg is not a split phase, it's just 240 volts.

It's getting late (even here), so maybe we'll just agree to disagree. But I will think about it some more.

Take care, my friend. Don't freeze out there tomorrow (eheh - take lots of warm up breaks). Sorry this got longer than I had intended.

Radar

Redsy - I read it that way (80KVA total 3Ø load) at first and had to erase my first post and refigure. It could easily be that, or you could have, say, an 80KW heater connected to each of the 3 pairs of terminals, which is how I've been looking at this.